A wedge Y with mass of 10 kg and all… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Mechanics

2026mcqhard

A wedge $Y$ with mass of $10$ kg and all frictionless surfaces and the inclined surface making $37°$ with horizontal. A block $X$ with mass $2$ kg is placed at the highest point of the wedge as shown in figure is at rest. At $t = 0$ wedge $(Y)$ is pulled toward right with constant force $(f)$ of $24$ N. Taking the block $X$ at rest at $t = 0$ , the time taken by it to slide down $8.8$ m on the slope, while $Y$ is on the move, is _____ s.

(take $\tan(37°) = 3/4$ and $g = 10$ m/s $^2$ )

JEE Main 2026 Physics, Laws of Motion — question figure

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$2$
B
$4$
C
$\sqrt{2}$
D
$2\sqrt{2}$

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Solution

Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally):

$A = \dfrac{f}{M + m} = \dfrac{24}{10 + 2} = 2\text{ m/s}^2$

Working in the non-inertial frame of the wedge, the block experiences a pseudo force $mA$ directed opposite to the wedge's acceleration. The components along the incline are:

Component of gravity down the incline: $g\sin 37^\circ$

Component of pseudo force up the incline: $A\cos 37^\circ$

Net relative acceleration of the block down the incline:

$a_{\text{rel}} = g\sin 37^\circ - A\cos 37^\circ$

Using $\sin 37^\circ = 0.6$ and $\cos 37^\circ = 0.8$ :

$a_{\text{rel}} = 10(0.6) - 2(0.8) = 6 - 1.6 = 4.4\text{ m/s}^2$

Time taken to slide a distance of $8.8\text{ m}$ along the incline, starting from rest:

$s = \dfrac{1}{2} a_{\text{rel}}\, t^2$

$8.8 = \dfrac{1}{2}(4.4)\,t^2$

$8.8 = 2.2\,t^2$

$t^2 = 4 \Rightarrow t = 2\text{ s}$

Hence, the required time is $2\text{ s}$ .

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